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Further Mathematics

Table of Contents
Question 1........................................................................................................................................1
Question 2........................................................................................................................................2
Question 3........................................................................................................................................2
Question 4........................................................................................................................................4
Question 5........................................................................................................................................5
Question 6........................................................................................................................................6
Question 7........................................................................................................................................6
Question 8........................................................................................................................................6
Question 9........................................................................................................................................6

Question 1.
(i) x2 – x – 2 = 0
Solution –
Solving equation By graphical method –
so, x = -1, 2
(ii) x3 – 8x + 5 – 3/x = 0
Solution –
x4 – 8x2 + 5x – 3 = 0
so, x = 3, -1
1

Question 2
Given,
initial velocity (u) = 15m/s
Motion of object –
ds = u – gt
dt
where, g = 9.8 m/sec2
Height of object after 2 sec, if s = 0 when = 0
Integrate the given equation as –
∫ds = ∫(u – gt). dt
s = ut – gt2
putting all the values,
s = 15 x 2 – 9.8 x (2)2
= 30 – 39.2
= 9.2m
Question 3
Given
Movement of body in straight line given as –
d2s + a2s = 0
dt2
where, a is constant and s = c and ds/dt = 0 at t = 2π/a
Integrate the given equation as –
d2s = - a2s
dt2
∫d2s = ∫a2s
dt2
ds = a2s t + C
dt
taking, a as constant and s = c and ds/dt = 0 at t = 2π/a
a2 c (2π/a) + C = 0
ac. 2π + C= 0
C = 2πac
then,
2

ds = a2s t + 2πac
dt
3

Question 4
Using trapezoidal rule and Simpson’s rule
i) ∫30 √ (1 + x4). dx
Solution:
Δx = b – a
n
here, a = 0, b = 3, n = 8
= 3 – 0 = 3
8 8
so, intervals – x = {0, 3/8, 3/4, 9/8, 3/2, 15/8, 9/4, 21/8, 3}
so, f(x0) = √ (1 + (0)4) = 1
f(x1) = √ (1 + (3/8)4) = 1.009
f(x2) = √ (1 + (3/4)4) = 1.147
f(x3) = √ (1 + (9/8)4) = 1.613
f(x4) = √ (1 + (3/2)4) = 2.462
f(x5) = √ (1 + (15/8)4) = 3.655
f(x6) = √ (1 + (9/4)4) = 5.160
f(x7) = √ (1 + (21/8)4) = 6.962
f(x8) = √ (1 + (3)4) = 9.055
Trapezoidal rule
≈ Δx/2 [f(x0) + 2. f(x1) + 2. f(x2) + 2. f(x3) + 2. f(x4) + 2. f(x5) + 2. f(x6) + 2. f(x7) + f(x8)]
≈ 3/16 [1 + 2x1.009 + 2x1.147+ 2x1.613+ 2x2.462 + 2x3.655 + 2x5.160 + 2x6.962 + 9.055]
≈ 8.084
Simpson’s rule
≈ Δx/3 [f(x0) + 4. f(x1) + 2. f(x2) + 4. f(x3) + 2. f(x4) + 4. f(x5) + 2. f(x6) + 4. f(x7) + f(x8)]
≈ 1/8 [1 + 4x1.009 + 2x1.147 + 4x1.613 + 2x2.462 + 4x3.655 + 2x5.160 + 4x6.962 + 9.005]
≈ 10.462
ii) ∫31 2ln3x dx
Solution:
4

Δx = b – a
n
here, a = 0, b = 3, n = 8
= 3 – 1 = 1
8 4
so, intervals – x = {1, 5/4 3/2, 7/4, 2, 9/4, 5/2, 11/4, 3}
so, f(x0) = 2ln3(1) = 2.197
f(x1) = 2ln3(5/4) = 2.643
f(x2) = 2ln3(3/2) = 3.008
f(x3) = 2ln3(7/4) = 3.316
f(x4) = 2ln3(2) = 3.583
f(x5) = 2ln3(9/4) = 3.819
f(x6) = 2ln3(5/2) = -1.248
f(x7) = 2ln3(11/4) = 4.220
f(x8) = 2ln3(3) = 4.394
Trapezoidal rule
≈ Δx/2 [f(x0) + 2. f(x1) + 2. f(x2) + 2. f(x3) + 2. f(x4) + 2. f(x5) + 2. f(x6) + 2. f(x7) + f(x8)]
≈ 1/8 [2.197 + 2x2.643 + 2x3.008 + 2x3.316+ 2x3.583+ 2x3.819 + 2x(-1.248) + 2x4.220 +
4.394]
≈ 5.659
Simpson’s rule
≈ Δx/3 [f(x0) + 4. f(x1) + 2. f(x2) + 4. f(x3) + 2. f(x4) + 4. f(x5) + 2. f(x6) + 4. f(x7) + f(x8)]
≈ 1/12 [2.197 + 4x2.643 + 2x3.008 + 4x3.316+ 2x3.583+ 4x3.819 + 2x(-1.248) + 4x4.220 +
4.394]
≈ 6.102
5

Question 5
Given equation –
CR dV + V = E
dt
(a) Solve equation for V at t = 0 and V = 0
Solution –
CR dV + V = E
dt
dV = 1 . dt
(E – V) CR
ln(E – V) = -t/CR + k
E – V = ke(-t/CR)
V = E - ke(-t/CR)
at initial condition, at V = 0 and t = 0
k = E
therefore,
V = E (1 – e(-t/CR))
(b) Calculate V when E = 25V, C = 20 x 10-6 F, R = 200 x 103Ω and t = 3.0s
Using, above equation –
V = E (1 – e(-t/CR))
= 25 (1 – e^(-3/20 x 10-6 . 200 x 103))
≈ 13.192
6

Question 6
Oscillations of a damped mechanical system –
2 d2y + 6dy + 4.5 y = 0
dt2 dt
displacement expression using Laplace Transforms –
at t = 0, y = 0 and dy/dt = 4
Solution –
2 d2y + 6dy + 4.5 y = 0
dt2 dt
Divide the given equation by 2,
d2y + 3dy + 2.25 y = 0
dt2 dt
Now, take the Laplace transform on both side –
Ĺ d2y + 3 Ĺdy + 2.25 Ĺy = Ĺ(0)
dt2 dt
solving it –
f(s) = 4 .
s + 6s +2.25
7

Question 7
Roots of equation using two different iterative techniques
x4 – 3x 3 + 7x = 12
Solution
Regula Falsi method
let y(x) = x4 – 3x 3 + 7x – 12
y(2) = -6
y(3) = 9
so, root of this equation lies between 2 and 3
Then, first approx.
x1= a f(b) – b f(a) /f(b) – f(a)
= 2 x (9) – 3 x (-6)/ 9 – (-6)
= 2.4
f(2.4) = (2.4)4 – 3 (2.4)3 + 7 (2.4) – 12
= 33.1776 – 41.472 + 16.8 – 12
= -3.5
so, root lies between 2.4 and 3
Then, second approx.
x2 = a f(b) – b f(a) /f(b) – f(a)
= 2.4 x (9) – 3 x (-3.5)/ 9 – (-3.5)
= 2.5
f(2.5) = (2.5)4 – 3 (2.5)3 + 7 (2.5) – 12
= 39.0625 – 46.875 + 17.5 – 12
= -2.3
so, root lies between 2.5 and 3
Then, third approx.
x3 = a f(b) – b f(a) /f(b) – f(a)
= 2.5 x (9) – 3 x (-2.3)/ 9 – (-2.3)
= 2.6
so, root of the given equation is 2.5
8

Newton’s iterative method –
let y(x) = x4 – 3x 3 + 7x – 12
y(2) = -6
y(3) = 9
so, root of this equation lies between 2 and 3
First approximation
x1 = (a+b)/2
= (2+3)/2 = 2.25
so, f(2.25) = (2.25)4 – 3 (2.25)3 + 7 (2.25) – 12
= 25.6289062 – 34.171875 + 15.75 – 12
= -4.8
Hence, root lies between 2.25 and 3
Second approximation
x2 = (2.25+3)/2
= 2.625
so, f(2.625) = (2.625)4 – 3 (2.625)3 + 7 (2.625) – 12
= -0.4
Hence, root lies between 2.625 and 3
Third approximation
x3 = (2.625+3)/2
= 2.8
so, f(2.8) = (2.8)4 – 3 (2.8)3 + 7 (2.8) – 12
= 3.2 which is positive
Hence, root of the equation is 2.8
9

Question 9
d2ϴ + 4 dϴ + 4ϴ = 8
dt2 dt
solve at t = 0, ϴ = dϴ/dt = 2
Solution –
Since the general solution of
(D2 + 4D + 4) ϴ = 8
substitute m for D
m2 + 4m + 4 = 8
m2 + 4m – 4 = 0
or, m = -2 ±2√ 2
As general solution of differential equation when roots are real
ϴ = (At + B)e^(-2 ±2√ 2)t
at t = 0 and ϴ = 0
2 = (A x 0 +B)e0
B = 2
while,
dϴ/dt = (At + B) (-2 ±2√ 2). e(-2 ±2√ 2)t + Ae(-2 ±2√ 2)t
2 = (0 + B) (-2 ±2√ 2). e0 + A. e0
A = 4√ 2
So, ϴ = (4√2t + 2)e^(-2 ±2√ 2)t
10